Question

The length and breadth of a rectangular sheet are 16. 2 cm and 10. 1 cm, respectively. The area of the sheet in appropriate significant figures and errors is

• A. $163.6\underset{–}{+}2.6c{m}^2$
• B. $164\underset{–}{+}3c{m}^2$
• C. $163.62\underset{–}{+}3c{m}^2$
• D. $163.62\underset{–}{+}2.6c{m}^2$

Answer 1

The correct option is B $164\underset{–}{+}3c{m}^2$

Step 1: Find the area of rectangular sheet.

Formula Used: $A=l\times b$
Given:
Length of sheet $\left(l\right)=16.2cm$
Breadth of sheet $\left(b\right)=10.1cm$
As we know,
Area of rectangular sheet(A)
= $Length\left(l\right)\times breadth\left(b\right)$
Therefore,
Area of rectangular sheet(A)
= $16.2cm\times 10.1cm$
Area of rectangular sheet (A) = $163.62c{m}^2$

Step 2: Find the error in area of rectangular sheet.

Error in product of quantities:

Suppose $x=a\times b$
Let $\Delta a$ = absolute error in measurement of a,
$\Delta b$ = absolute error in measurement of b,
$\Delta x$ = absolute error in calculation of x, i.e., product of a and b
The maximum fractional error in x is $\frac{\Delta x}{x}=\underset{–}{+}(\frac{\Delta a}{a}+\frac{\Delta b}{b})$ According to the problem,
length $l=\left(16.2\underset{–}{+}0.1\right)cm$

breadth $b=\left(10.1\underset{–}{+}0.1\right)cm$

As we know, Area of rectangular sheet = $163.62c{m}^2$ As per the rule, area will have only three significant figures and error will have only one significant figure. Rounding off we get, area A = $164c{m}^2$
If $\Delta A$ is error in the area, then relative error is calculated as $\frac{\Delta A}{A}$

$\frac{\Delta A}{A}$ = $\frac{\Delta l}{l}+\frac{\Delta b}{b}$

$\frac{\Delta A}{A}$ = $\frac{0.1cm}{16.2cm}+\frac{0.1cm}{10.1cm}$

$\frac{\Delta A}{A}$ = $\frac{1.01+1.62}{16.2\times 10.1}$ = $\frac{2.63}{163.62}$

$\Delta A=A\times \frac{2.63}{163.62}c{m}^2$

$\Delta A=162.62\times \frac{2.63}{163.62}=2.63c{m}^2$

$\Delta A=3c{m}^2$ (By rounding off to one significant figure)

Therefore, area of rectangular sheet in significant figure and error is given by:

Area $A=A\underset{–}{+}\Delta A$

A = $\left(164\underset{–}{+}3\right)c{m}^2$

Final answer: A = $\left(164\underset{–}{+}3\right)c{m}^2$